IMO Hk TST, 1997

(a) We have
$$\begin{aligned} & \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \left(\frac{1}{a} - \frac{1}{b}\right)^2 - \left(\frac{1}{b} - \frac{1}{c}\right)^2 - \left(\frac{1}{c} - \frac{1}{a}\right)^2 \\ &= 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right) \\ &= \left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) - \frac{1}{2}\left[\left(\frac{1}{a} - \frac{1}{b}\right)^2 + \left(\frac{1}{b} - \frac{1}{c}\right)^2 + \left(\frac{1}{c} - \frac{1}{a}\right)^2\right] \\ &\le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{abc} \\ &= \frac{a+b+c}{4RS} = \frac{\sin A + \sin B + \sin C}{2S}, \end{aligned}$$
where we have used the fact $S = \frac{abc}{4R}$ and the extended sine law. Now, note that $f(x) = \sin x$ is a concave function on $(0, \pi)$ since $f''(x) = -\sin x < 0$. Therefore, by Jensen's inequality, we have
$$\frac{\sin A + \sin B + \sin C}{2S} \le \frac{3}{2S} \sin \frac{A+B+C}{3} = \frac{3\sqrt{3}}{4S}.$$
The result follows readily. Equality holds when $a = b = c$.

(b) Let $r$ and $s$ be the inradius and the semiperimeter of $\triangle ABC$. Then we have
$$\tan \frac{A}{2} = \frac{r}{s-a}$$
etc. Also, using $S = rs = \sqrt{s(s-a)(s-b)(s-c)}$, we have
$$\begin{align*} & \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} \le \frac{9R^2}{4S} \\ \Leftrightarrow & \qquad \frac{r}{s-a} + \frac{r}{s-b} + \frac{r}{s-c} \le \frac{9R^2}{4S} \\ \Leftrightarrow & \frac{rs[(s-a)(s-b) + (s-b)(s-c) + (s-c)(s-a)]}{s(s-a)(s-b)(s-c)} \le \frac{9R^2}{4S} \\ \Leftrightarrow & 3s^2 - 2(a+b+c)s + (ab+bc+ca) \le \frac{9R^2}{4} \\ \Leftrightarrow & -(a+b+c)^2 + 4(ab+bc+ca) \le 9R^2. \end{align*}$$
Recall that
$$9R^2 - (a^2 + b^2 + c^2) = OH^2 \ge 0,$$
where $O$ and $H$ are the circumcentre and orthocentre of $\triangle ABC$ respectively.
Therefore, it suffices to prove
$$-(a + b + c)^2 + 4(ab + bc + ca) \le a^2 + b^2 + c^2.$$
This is equivalent to
$$ab + bc + ca \le a^2 + b^2 + c^2,$$
which is a well-known inequality. Thus, the inequality is proven. Equality holds when $\triangle ABC$ is equilateral.